Chapter 5.9, Problem 21E

(a)

To determine

To find:

The value of the given Simpson’s rule.

Answer to Problem 21E

  $\left|{\mathrm{f}}^{″}\left(\mathrm{x}\right)\right|=\left(\left|{\sin }^2\mathrm{x}-\cos \mathrm{x}\right|\right){\mathrm{e}}^{\cos \mathrm{x}}$

Explanation of Solution

Given:

The equation

  .${\displaystyle \underset{0}{\overset{2\mathrm{\pi }}{\int }}{\mathrm{e}}^{\cos \mathrm{x}}\mathrm{d}\mathrm{x}}$

Concept used:

Simpson’s error formula for $\mathrm{n}$

  $\left|{\mathrm{E}}_{\mathrm{s}}\right|\le \frac{\mathrm{K}{\left(\mathrm{b}-\mathrm{a}\right)}^5}{180{\mathrm{n}}^4}$

Calculation:

The function

  $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\cos \mathrm{x}}$

Differentiating with respect to $\mathrm{x}$

  $\begin{array}{l}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=-\sin \mathrm{x}{\mathrm{e}}^{\cos \mathrm{x}}\\ {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=-\sin \mathrm{x}\mathrm{f}\left(\mathrm{x}\right)\\ {\mathrm{f}}^{″}\left(\mathrm{x}\right)=-\cos \mathrm{x}\mathrm{f}\left(\mathrm{x}\right)-\sin \mathrm{x}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\\ {\mathrm{f}}^{″}\left(\mathrm{x}\right)=\left({\sin }^2\mathrm{x}-\cos \mathrm{x}\right){\mathrm{e}}^{\cos \mathrm{x}}\\ \left|{\mathrm{f}}^{″}\left(\mathrm{x}\right)\right|=\left(\left|{\sin }^2\mathrm{x}-\cos \mathrm{x}\right|\right){\mathrm{e}}^{\cos \mathrm{x}}\end{array}$

Draw the graph

  

(b)

To determine

To find:

The value of the given mid-point rule.

Answer to Problem 21E

  ${\mathrm{M}}_{10}=7.954927$

Explanation of Solution

Given:

The equation

  .${\displaystyle \underset{0}{\overset{2\mathrm{\pi }}{\int }}{\mathrm{e}}^{\cos \mathrm{x}}\mathrm{d}\mathrm{x}}$

Concept used:

Simpson’s error formula for $\mathrm{n}$

  $\left|{\mathrm{E}}_{\mathrm{s}}\right|\le \frac{\mathrm{K}{\left(\mathrm{b}-\mathrm{a}\right)}^5}{180{\mathrm{n}}^4}$

Calculation:

Draw the table

$\mathrm{x}$	$\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$	$\raisebox{1ex}{$3\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$	$\raisebox{1ex}{$5\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$	$\raisebox{1ex}{$7\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$	$\raisebox{1ex}{$9\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$	$\raisebox{1ex}{$11\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$	$\raisebox{1ex}{$13\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$	$\raisebox{1ex}{$15\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$	$\raisebox{1ex}{$17\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$	$\raisebox{1ex}{$19\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$
${\mathrm{e}}^{\cos \mathrm{x}}$	$2.5884429$	$1.7999975$	$1$	$0.55555634$	$0.38633264$	$0.38633264$	$0.55555634$	$1$	$1.7999975$	$2.5884429$

  .${\displaystyle \underset{0}{\overset{2\mathrm{\pi }}{\int }}{\mathrm{e}}^{\cos \mathrm{x}}\mathrm{d}\mathrm{x}}$

Using midpoint rule

  ${\displaystyle \underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}}=\mathrm{\Delta }\mathrm{x}\left[\mathrm{f}\left(\overline{{\mathrm{x}}_1}\right)+\mathrm{f}\left(\overline{{\mathrm{x}}_2}\right)+\mathrm{.............}\mathrm{f}\left(\overline{{\mathrm{x}}_{\mathrm{n}}}\right)\right]$

  $\mathrm{\Delta }\mathrm{x}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}=\frac{\mathrm{\pi }}{5}$

  ${\mathrm{x}}_{\mathrm{i}}=\frac{1}{2}\left({\mathrm{x}}_{\mathrm{i}-1}+{\mathrm{x}}_{\mathrm{i}}\right)$

Using the above formula

  $\begin{array}{l}{\mathrm{M}}_{10}=\mathrm{\Delta }\mathrm{x}\left[\mathrm{f}\left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\right)+\mathrm{f}\left(\raisebox{1ex}{$3\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\right)+\mathrm{f}\left(\raisebox{1ex}{$5\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\right)+\mathrm{f}\left(\raisebox{1ex}{$7\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\right)+\mathrm{f}\left(\raisebox{1ex}{$9\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\right)+\mathrm{f}\left(\raisebox{1ex}{$11\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\right)+\mathrm{f}\left(\raisebox{1ex}{$13\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\right)+\mathrm{f}\left(\raisebox{1ex}{$15\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\right)+\mathrm{f}\left(\raisebox{1ex}{$17\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\right)+\mathrm{f}\left(\raisebox{1ex}{$19\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\right)\right]\\ {\mathrm{M}}_{10}=7.954927\end{array}$

(c)

To determine

To find:

The value of the given mid-point error rule.

Answer to Problem 21E

  $\left|{\mathrm{E}}_{\mathrm{M}}\right|\le \approx 0.2894$

Explanation of Solution

Given:

The equation

  .${\displaystyle \underset{0}{\overset{2\mathrm{\pi }}{\int }}{\mathrm{e}}^{\cos \mathrm{x}}\mathrm{d}\mathrm{x}}$

Concept used:

Simpson’s error formula for $\mathrm{n}$

  $\left|{\mathrm{E}}_{\mathrm{s}}\right|\le \frac{\mathrm{K}{\left(\mathrm{b}-\mathrm{a}\right)}^5}{180{\mathrm{n}}^4}$

Calculation:

The function

  $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\cos \mathrm{x}}$

Differentiating with respect to $\mathrm{x}$

  $\begin{array}{l}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=-\sin \mathrm{x}{\mathrm{e}}^{\cos \mathrm{x}}\\ {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=-\sin \mathrm{x}\mathrm{f}\left(\mathrm{x}\right)\\ {\mathrm{f}}^{″}\left(\mathrm{x}\right)=-\cos \mathrm{x}\mathrm{f}\left(\mathrm{x}\right)-\sin \mathrm{x}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\\ {\mathrm{f}}^{″}\left(\mathrm{x}\right)=\left({\sin }^2\mathrm{x}-\cos \mathrm{x}\right){\mathrm{e}}^{\cos \mathrm{x}}\\ \left|{\mathrm{f}}^{″}\left(\mathrm{x}\right)\right|=\left(\left|{\sin }^2\mathrm{x}-\cos \mathrm{x}\right|\right){\mathrm{e}}^{\cos \mathrm{x}}\end{array}$

  $\begin{array}{l}\mathrm{k}=2.8,\mathrm{a}=0,\mathrm{b}=2\mathrm{\pi },\mathrm{n}=10\\ \left|{\mathrm{E}}_{\mathrm{M}}\right|\le \frac{\mathrm{K}{\left(\mathrm{b}-\mathrm{a}\right)}^3}{24{\mathrm{n}}^2}\\ \left|{\mathrm{E}}_{\mathrm{M}}\right|\le \frac{2.8{\left(2\mathrm{\pi }-0\right)}^3}{\left(24\right)6^2}\\ \left|{\mathrm{E}}_{\mathrm{M}}\right|\le \approx 0.2894\end{array}$

(d)

To determine

To find:

The built-in numerical integration capability of CAS to approximate I.

Answer to Problem 21E

  ${\displaystyle \underset{0}{\overset{2\mathrm{\pi }}{\int }}{\mathrm{e}}^{\cos \mathrm{x}}\mathrm{d}\mathrm{x}}\approx 7.95493$

Explanation of Solution

Given:

The equation

  .${\displaystyle \underset{0}{\overset{2\mathrm{\pi }}{\int }}{\mathrm{e}}^{\cos \mathrm{x}}\mathrm{d}\mathrm{x}}$

Concept used:

Simpson’s error formula for $\mathrm{n}$

  $\left|{\mathrm{E}}_{\mathrm{s}}\right|\le \frac{\mathrm{K}{\left(\mathrm{b}-\mathrm{a}\right)}^5}{180{\mathrm{n}}^4}$

Calculation:

The function

  $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\cos \mathrm{x}}$

Differentiating with respect to $\mathrm{x}$

  $\begin{array}{l}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=-\sin \mathrm{x}{\mathrm{e}}^{\cos \mathrm{x}}\\ {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=-\sin \mathrm{x}\mathrm{f}\left(\mathrm{x}\right)\\ {\mathrm{f}}^{″}\left(\mathrm{x}\right)=-\cos \mathrm{x}\mathrm{f}\left(\mathrm{x}\right)-\sin \mathrm{x}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\\ {\mathrm{f}}^{″}\left(\mathrm{x}\right)=\left({\sin }^2\mathrm{x}-\cos \mathrm{x}\right){\mathrm{e}}^{\cos \mathrm{x}}\\ \left|{\mathrm{f}}^{″}\left(\mathrm{x}\right)\right|=\left(\left|{\sin }^2\mathrm{x}-\cos \mathrm{x}\right|\right){\mathrm{e}}^{\cos \mathrm{x}}\end{array}$

  $\begin{array}{l}\mathrm{k}=2.8,\mathrm{a}=0,\mathrm{b}=2\mathrm{\pi },\mathrm{n}=10\\ \left|{\mathrm{E}}_{\mathrm{M}}\right|\le \frac{\mathrm{K}{\left(\mathrm{b}-\mathrm{a}\right)}^3}{24{\mathrm{n}}^2}\\ \left|{\mathrm{E}}_{\mathrm{M}}\right|\le \frac{2.8{\left(2\mathrm{\pi }-0\right)}^3}{\left(24\right)6^2}\\ \left|{\mathrm{E}}_{\mathrm{M}}\right|\le \approx 0.2894\end{array}$

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{2\mathrm{\pi }}{\int }}{\mathrm{e}}^{\cos \mathrm{x}}\mathrm{d}\mathrm{x}}=2\mathrm{\pi }{\mathrm{I}}_0\left(1\right)\\ {\displaystyle \underset{0}{\overset{2\mathrm{\pi }}{\int }}{\mathrm{e}}^{\cos \mathrm{x}}\mathrm{d}\mathrm{x}}\approx 7.95493\end{array}$

(e)

To determine

To find:

The value of the given real error rule.

Answer to Problem 21E

  $3\times {10}^{-6}$

Explanation of Solution

Given:

The equation

  .${\displaystyle \underset{0}{\overset{2\mathrm{\pi }}{\int }}{\mathrm{e}}^{\cos \mathrm{x}}\mathrm{d}\mathrm{x}}$

Concept used:

Simpson’s error formula for $\mathrm{n}$

  $\left|{\mathrm{E}}_{\mathrm{s}}\right|\le \frac{\mathrm{K}{\left(\mathrm{b}-\mathrm{a}\right)}^5}{180{\mathrm{n}}^4}$

Calculation:

The function

  $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\cos \mathrm{x}}$

Differentiating with respect to $\mathrm{x}$

  $\begin{array}{l}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=-\sin \mathrm{x}{\mathrm{e}}^{\cos \mathrm{x}}\\ {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=-\sin \mathrm{x}\mathrm{f}\left(\mathrm{x}\right)\\ {\mathrm{f}}^{″}\left(\mathrm{x}\right)=-\cos \mathrm{x}\mathrm{f}\left(\mathrm{x}\right)-\sin \mathrm{x}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\\ {\mathrm{f}}^{″}\left(\mathrm{x}\right)=\left({\sin }^2\mathrm{x}-\cos \mathrm{x}\right){\mathrm{e}}^{\cos \mathrm{x}}\\ \left|{\mathrm{f}}^{″}\left(\mathrm{x}\right)\right|=\left(\left|{\sin }^2\mathrm{x}-\cos \mathrm{x}\right|\right){\mathrm{e}}^{\cos \mathrm{x}}\end{array}$

  $\begin{array}{l}\mathrm{k}=2.8,\mathrm{a}=0,\mathrm{b}=2\mathrm{\pi },\mathrm{n}=10\\ \left|{\mathrm{E}}_{\mathrm{M}}\right|\le \frac{\mathrm{K}{\left(\mathrm{b}-\mathrm{a}\right)}^3}{24{\mathrm{n}}^2}\\ \left|{\mathrm{E}}_{\mathrm{M}}\right|\le \frac{2.8{\left(2\mathrm{\pi }-0\right)}^3}{\left(24\right)6^2}\\ \left|{\mathrm{E}}_{\mathrm{M}}\right|\le \approx 0.2894\end{array}$

  ${\displaystyle \underset{0}{\overset{2\mathrm{\pi }}{\int }}{\mathrm{e}}^{\cos \mathrm{x}}\mathrm{d}\mathrm{x}}\approx 7.95493$

The real error

  $\begin{array}{l}=7.95493-7.954927\\ =3\times {10}^{-6}\end{array}$

(f)

To determine

To find:

The value of the $\left|{\mathrm{f}}^4\left(\mathrm{x}\right)\right|$ given rule.

Answer to Problem 21E

  $\mathrm{K}\approx 10.8731273138$

Explanation of Solution

Given:

The equation

  .${\displaystyle \underset{0}{\overset{2\mathrm{\pi }}{\int }}{\mathrm{e}}^{\cos \mathrm{x}}\mathrm{d}\mathrm{x}}$

Concept used:

Simpson’s error formula for $\mathrm{n}$

  $\left|{\mathrm{E}}_{\mathrm{s}}\right|\le \frac{\mathrm{K}{\left(\mathrm{b}-\mathrm{a}\right)}^5}{180{\mathrm{n}}^4}$

Calculation:

The function

  $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\cos \mathrm{x}}$

Differentiating with respect to $\mathrm{x}$

  $\begin{array}{l}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=-\sin \mathrm{x}{\mathrm{e}}^{\cos \mathrm{x}}\\ {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=-\sin \mathrm{x}\mathrm{f}\left(\mathrm{x}\right)\\ {\mathrm{f}}^{″}\left(\mathrm{x}\right)=-\cos \mathrm{x}\mathrm{f}\left(\mathrm{x}\right)-\sin \mathrm{x}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\\ {\mathrm{f}}^{″}\left(\mathrm{x}\right)=\left({\sin }^2\mathrm{x}-\cos \mathrm{x}\right){\mathrm{e}}^{\cos \mathrm{x}}\\ \left|{\mathrm{f}}^{″}\left(\mathrm{x}\right)\right|=\left(\left|{\sin }^2\mathrm{x}-\cos \mathrm{x}\right|\right){\mathrm{e}}^{\cos \mathrm{x}}\end{array}$

  ${\mathrm{f}}^4\left(\mathrm{x}\right)=\left({\sin }^4\mathrm{x}-4{\sin }^2\mathrm{x}+3{\cos }^2\mathrm{x}+\cos \mathrm{x}-6{\sin }^2\mathrm{x}\cos \mathrm{x}\right){\mathrm{e}}^{\cos \mathrm{x}}$

Putting $\mathrm{x}=0$

  $\begin{array}{l}\mathrm{K}={\mathrm{f}}^4\left(0\right)=\left({\sin }^{4}0-4{\sin }^{2}0+3{\cos }^{2}0+\cos 0-6{\sin }^{2}0\cos 0\right){\mathrm{e}}^{\cos 0}\\ \mathrm{K}=\mathrm{e}\left(0-0+3+1-0\right)\\ \mathrm{K}=4\mathrm{e}\\ \mathrm{K}\approx 10.8731273138\end{array}$

Draw the graph

  

(g)

To determine

To find:

The value of the given Simpson’s rule.

Answer to Problem 21E

  ${\mathrm{S}}_{10}=7.953789$

Explanation of Solution

Given:

The equation

  .${\displaystyle \underset{0}{\overset{2\mathrm{\pi }}{\int }}{\mathrm{e}}^{\cos \mathrm{x}}\mathrm{d}\mathrm{x}}$

Concept used:

Simpson’s error formula for $\mathrm{n}$

  $\left|{\mathrm{E}}_{\mathrm{s}}\right|\le \frac{\mathrm{K}{\left(\mathrm{b}-\mathrm{a}\right)}^5}{180{\mathrm{n}}^4}$

Calculation:

Draw the table

$\mathrm{x}$	$\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$	$\raisebox{1ex}{$3\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$	$\raisebox{1ex}{$5\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$	$\raisebox{1ex}{$7\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$	$\raisebox{1ex}{$9\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$	$\raisebox{1ex}{$11\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$	$\raisebox{1ex}{$13\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$	$\raisebox{1ex}{$15\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$	$\raisebox{1ex}{$17\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$	$\raisebox{1ex}{$19\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$
${\mathrm{e}}^{\cos \mathrm{x}}$	$2.5884429$	$1.7999975$	$1$	$0.55555634$	$0.38633264$	$0.38633264$	$0.55555634$	$1$	$1.7999975$	$2.5884429$

  .${\displaystyle \underset{0}{\overset{2\mathrm{\pi }}{\int }}{\mathrm{e}}^{\cos \mathrm{x}}\mathrm{d}\mathrm{x}}$

Using Simpson’s rule

  ${\displaystyle \underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}\mathrm{x}}=\mathrm{\Delta }\mathrm{x}\left[\mathrm{f}\left(\overline{{\mathrm{x}}_1}\right)+\mathrm{f}\left(\overline{{\mathrm{x}}_2}\right)+\mathrm{.............}\mathrm{f}\left(\overline{{\mathrm{x}}_{\mathrm{n}}}\right)\right]$

  $\mathrm{\Delta }\mathrm{x}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}=\frac{\mathrm{\pi }}{5}$

Using the above formula

  $\begin{array}{l}{\mathrm{S}}_{10}=\mathrm{\Delta }\mathrm{x}\left[\mathrm{f}\left(0\right)+4\mathrm{f}\left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)+2\mathrm{f}\left(\raisebox{1ex}{$2\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)+4\mathrm{f}\left(\raisebox{1ex}{$3\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)+2\mathrm{f}\left(\raisebox{1ex}{$4\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)+4\mathrm{f}\left(\mathrm{\pi }\right)+2\mathrm{f}\left(\raisebox{1ex}{$6\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)+4\mathrm{f}\left(\raisebox{1ex}{$7\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)+2\mathrm{f}\left(\raisebox{1ex}{$8\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)+4\mathrm{f}\left(\raisebox{1ex}{$9\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)+\mathrm{f}\left(2\mathrm{\pi }\right)\right]\\ {\mathrm{S}}_{10}=7.953789\end{array}$

(h)

To determine

To find:

The value of the error of the given rule.

Answer to Problem 21E

  $\left|{\mathrm{E}}_{\mathrm{S}}\right|\approx 0.059299814$

Explanation of Solution

Given:

The equation

  .${\displaystyle \underset{0}{\overset{2\mathrm{\pi }}{\int }}{\mathrm{e}}^{\cos \mathrm{x}}\mathrm{d}\mathrm{x}}$

Concept used:

Simpson’s error formula for $\mathrm{n}$

  $\left|{\mathrm{E}}_{\mathrm{s}}\right|\le \frac{\mathrm{K}{\left(\mathrm{b}-\mathrm{a}\right)}^5}{180{\mathrm{n}}^4}$

Calculation:

The function

  $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\cos \mathrm{x}}$

Differentiating with respect to $\mathrm{x}$

  $\begin{array}{l}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=-\sin \mathrm{x}{\mathrm{e}}^{\cos \mathrm{x}}\\ {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=-\sin \mathrm{x}\mathrm{f}\left(\mathrm{x}\right)\\ {\mathrm{f}}^{″}\left(\mathrm{x}\right)=-\cos \mathrm{x}\mathrm{f}\left(\mathrm{x}\right)-\sin \mathrm{x}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\\ {\mathrm{f}}^{″}\left(\mathrm{x}\right)=\left({\sin }^2\mathrm{x}-\cos \mathrm{x}\right){\mathrm{e}}^{\cos \mathrm{x}}\\ \left|{\mathrm{f}}^{″}\left(\mathrm{x}\right)\right|=\left(\left|{\sin }^2\mathrm{x}-\cos \mathrm{x}\right|\right){\mathrm{e}}^{\cos \mathrm{x}}\end{array}$

  ${\mathrm{f}}^4\left(\mathrm{x}\right)=\left({\sin }^4\mathrm{x}-4{\sin }^2\mathrm{x}+3{\cos }^2\mathrm{x}+\cos \mathrm{x}-6{\sin }^2\mathrm{x}\cos \mathrm{x}\right){\mathrm{e}}^{\cos \mathrm{x}}$

Putting $\mathrm{x}=0$

  $\begin{array}{l}\mathrm{K}={\mathrm{f}}^4\left(0\right)=\left({\sin }^{4}0-4{\sin }^{2}0+3{\cos }^{2}0+\cos 0-6{\sin }^{2}0\cos 0\right){\mathrm{e}}^{\cos 0}\\ \mathrm{K}=\mathrm{e}\left(0-0+3+1-0\right)\\ \mathrm{K}=4\mathrm{e}\\ \mathrm{K}\approx 10.8731273138\end{array}$

  $\begin{array}{l}\mathrm{k}=4\mathrm{e},\mathrm{a}=0,\mathrm{b}=2\mathrm{\pi },\mathrm{n}=10\\ \left|{\mathrm{E}}_{\mathrm{S}}\right|\le \frac{\mathrm{K}{\left(\mathrm{b}-\mathrm{a}\right)}^5}{180{\mathrm{n}}^4}\\ \left|{\mathrm{E}}_{\mathrm{S}}\right|\le \frac{10.9{\left(2\mathrm{\pi }-0\right)}^5}{\left(180\right){10}^4}\\ \left|{\mathrm{E}}_{\mathrm{S}}\right|\approx 0.059299814\end{array}$

(i)

To determine

To find:

The actual value error compare with the error estimate of the given rule.

Answer to Problem 21E

  $0.00114$

Explanation of Solution

Given:

The equation

  .${\displaystyle \underset{0}{\overset{2\mathrm{\pi }}{\int }}{\mathrm{e}}^{\cos \mathrm{x}}\mathrm{d}\mathrm{x}}$

Concept used:

Simpson’s error formula for $\mathrm{n}$

  $\left|{\mathrm{E}}_{\mathrm{s}}\right|\le \frac{\mathrm{K}{\left(\mathrm{b}-\mathrm{a}\right)}^5}{180{\mathrm{n}}^4}$

Calculation:

The function

  $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\cos \mathrm{x}}$

Differentiating with respect to $\mathrm{x}$

  $\begin{array}{l}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=-\sin \mathrm{x}{\mathrm{e}}^{\cos \mathrm{x}}\\ {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=-\sin \mathrm{x}\mathrm{f}\left(\mathrm{x}\right)\\ {\mathrm{f}}^{″}\left(\mathrm{x}\right)=-\cos \mathrm{x}\mathrm{f}\left(\mathrm{x}\right)-\sin \mathrm{x}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\\ {\mathrm{f}}^{″}\left(\mathrm{x}\right)=\left({\sin }^2\mathrm{x}-\cos \mathrm{x}\right){\mathrm{e}}^{\cos \mathrm{x}}\\ \left|{\mathrm{f}}^{″}\left(\mathrm{x}\right)\right|=\left(\left|{\sin }^2\mathrm{x}-\cos \mathrm{x}\right|\right){\mathrm{e}}^{\cos \mathrm{x}}\end{array}$

  ${\mathrm{f}}^4\left(\mathrm{x}\right)=\left({\sin }^4\mathrm{x}-4{\sin }^2\mathrm{x}+3{\cos }^2\mathrm{x}+\cos \mathrm{x}-6{\sin }^2\mathrm{x}\cos \mathrm{x}\right){\mathrm{e}}^{\cos \mathrm{x}}$

Putting $\mathrm{x}=0$

  $\begin{array}{l}\mathrm{K}={\mathrm{f}}^4\left(0\right)=\left({\sin }^{4}0-4{\sin }^{2}0+3{\cos }^{2}0+\cos 0-6{\sin }^{2}0\cos 0\right){\mathrm{e}}^{\cos 0}\\ \mathrm{K}=\mathrm{e}\left(0-0+3+1-0\right)\\ \mathrm{K}=4\mathrm{e}\\ \mathrm{K}\approx 10.8731273138\end{array}$

  $\begin{array}{l}\mathrm{k}=4\mathrm{e},\mathrm{a}=0,\mathrm{b}=2\mathrm{\pi },\mathrm{n}=10\\ \left|{\mathrm{E}}_{\mathrm{S}}\right|\le \frac{\mathrm{K}{\left(\mathrm{b}-\mathrm{a}\right)}^5}{180{\mathrm{n}}^4}\\ \left|{\mathrm{E}}_{\mathrm{S}}\right|\le \frac{10.9{\left(2\mathrm{\pi }-0\right)}^5}{\left(180\right){10}^4}\\ \left|{\mathrm{E}}_{\mathrm{S}}\right|\le \approx 0.059299814\end{array}$

The actual error is

  $\begin{array}{l}=7.954926521-7.953789422\\ =0.00114\end{array}$

(j)

To determine

To find:

The size of the error in ${\mathrm{S}}_{\mathrm{n}}$ is less than $0.0001$ by using of the given rule.

Answer to Problem 21E

  $\mathrm{n}\ge 50$

Explanation of Solution

Given:

The equation

  .${\displaystyle \underset{0}{\overset{2\mathrm{\pi }}{\int }}{\mathrm{e}}^{\cos \mathrm{x}}\mathrm{d}\mathrm{x}}$

Concept used:

Simpson’s error formula for $\mathrm{n}$

  $\left|{\mathrm{E}}_{\mathrm{s}}\right|\le \frac{\mathrm{K}{\left(\mathrm{b}-\mathrm{a}\right)}^5}{180{\mathrm{n}}^4}$

Calculation:

The function

  $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\cos \mathrm{x}}$

Differentiating with respect to $\mathrm{x}$

  $\begin{array}{l}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=-\sin \mathrm{x}{\mathrm{e}}^{\cos \mathrm{x}}\\ {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=-\sin \mathrm{x}\mathrm{f}\left(\mathrm{x}\right)\\ {\mathrm{f}}^{″}\left(\mathrm{x}\right)=-\cos \mathrm{x}\mathrm{f}\left(\mathrm{x}\right)-\sin \mathrm{x}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\\ {\mathrm{f}}^{″}\left(\mathrm{x}\right)=\left({\sin }^2\mathrm{x}-\cos \mathrm{x}\right){\mathrm{e}}^{\cos \mathrm{x}}\\ \left|{\mathrm{f}}^{″}\left(\mathrm{x}\right)\right|=\left(\left|{\sin }^2\mathrm{x}-\cos \mathrm{x}\right|\right){\mathrm{e}}^{\cos \mathrm{x}}\end{array}$

  ${\mathrm{f}}^4\left(\mathrm{x}\right)=\left({\sin }^4\mathrm{x}-4{\sin }^2\mathrm{x}+3{\cos }^2\mathrm{x}+\cos \mathrm{x}-6{\sin }^2\mathrm{x}\cos \mathrm{x}\right){\mathrm{e}}^{\cos \mathrm{x}}$

Putting $\mathrm{x}=0$

  $\begin{array}{l}\mathrm{K}={\mathrm{f}}^4\left(0\right)=\left({\sin }^{4}0-4{\sin }^{2}0+3{\cos }^{2}0+\cos 0-6{\sin }^{2}0\cos 0\right){\mathrm{e}}^{\cos 0}\\ \mathrm{K}=\mathrm{e}\left(0-0+3+1-0\right)\\ \mathrm{K}=4\mathrm{e}\\ \mathrm{K}\approx 10.8731273138\end{array}$

  $\begin{array}{l}\mathrm{k}=4\mathrm{e},\mathrm{a}=0,\mathrm{b}=2\mathrm{\pi },\mathrm{n}=10\\ \left|{\mathrm{E}}_{\mathrm{S}}\right|\le 0.0001\\ \left|{\mathrm{E}}_{\mathrm{S}}\right|\le \frac{\mathrm{K}{\left(\mathrm{b}-\mathrm{a}\right)}^5}{180{\mathrm{n}}^4}\\ \left|{\mathrm{E}}_{\mathrm{S}}\right|\le \frac{4\mathrm{e}{\left(2\mathrm{\pi }-0\right)}^5}{\left(180\right)\times 0.0001}\le {\mathrm{n}}^4\\ {\mathrm{n}}^4\ge 5,915,362\\ \mathrm{n}\ge 49.3\\ \mathrm{n}\ge 50\end{array}$